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Jan 22, 2021 at 19:43 vote accept Roy
Jan 18, 2021 at 17:13 comment added Sébastien Loisel Also, from my point of view, there are two values on each edge. People often prefer to think of single-valued functions on edges by taking the average. Certainly, the continuity condition (2) will force single-valuedness. Thus, at convergence, you would be taking the average of two identical functions across that edge.
Jan 18, 2021 at 17:11 comment added Sébastien Loisel I'm not sure what you're trying to integrate over what edge. The flux across an edge can be recovered from (1) because the boundary terms enumerate all the edges. You might prefer to have the one edge integral on the left of the = in (1), and put the rest on the right side of the =. Note also that if you only want local information, you can restrict your attention to test functions $v(x)$ that are in the neighborhood of that edge; this removes a lot of faraway boundary terms.
Jan 12, 2021 at 9:51 comment added Roy Thanks, but what if we need to integrate over a line (not necessarily on the boundary)? So is the values on the edge an average of 2 elements sharing the edge?
Jan 8, 2021 at 19:58 history answered Sébastien Loisel CC BY-SA 4.0